Calculate the pH when 64.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

Question

Calculate the pH when 64.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

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Ngọc Hoa 5 months 2021-09-05T06:18:47+00:00 1 Answers 8 views 0

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    0
    2021-09-05T06:19:56+00:00

    Answer:

    The answer is “12.06

    Explanation:

    Given:

    M(HBrO) = 0.3\ M\\\\V(HBrO) = 20 \ mL\\\\M(KOH) = 0.15 \ M\\\\V(KOH) = 64 \ mL

    \to mol(HBrO) = M(HBrO)  \times  V(HBrO) = 0.3 M   \times 20 mL = 6 \ mmol\\\\\to mol(KOH) = M(KOH)  \times   V(KOH)= 0.15 M  \times  64 mL = 9.6 mmol

     
    6 mmol of both will react

    excess KOH remaining= 3.15 \ mmol

    Volume= 20 + 64 = 84 \ mL

    [OH^{-}] = \frac{ 9.6 \ mmol}{84\  mL} = 0.01142\ M

    use:

    pOH = -\log [OH^-]

            = -\log (1.142\times 10^{-2})\\\\= 1.94

    use:

    PH = 14 - pOH

           = 14 - 1.94\\\\= 12.06

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